Calculus

Applications of the Derivative

Applications of Derivative Logo

Critical Points

Solved Problems

Example 7.

Indicate all critical points of the function \[f\left( x \right) = \left| {{x^2} - 5} \right|.\]

Solution.

Find the roots of the function:

\[f\left( x \right) = 0,\;\; \Rightarrow \left| {{x^2} - 5} \right| = 0,\;\; \Rightarrow {x_{1,2}} = \pm \sqrt 5 .\]

The derivative does not exist at the corner points \(x = - \sqrt 5 \) and \(x = \sqrt 5 ,\) i.e. these points are critical.

In the interval \(\left[ { - \sqrt 5 ,\sqrt 5 } \right],\) the function is written as

\[f\left( x \right) = - \left( {{x^2} - 5} \right) = - {x^2} + 5.\]

Solving the equation \(f^\prime\left( c \right) = 0\) on this interval, we get one more critical point:

\[f^\prime\left( c \right) = 0,\;\; \Rightarrow - 2c = 0,\;\; \Rightarrow c = 0.\]

Hence, the function has three critical points:

\[{c_1} = - \sqrt 5,\;{c_2} = 0,\;{c_3} = \sqrt 5 .\]

Example 8.

Find all critical points of the function \[f\left( x \right) = \frac{x}{{\ln x}}.\]

Solution.

The domain of \(f\left( x \right)\) is determined by the conditions:

\[\left\{ \begin{array}{l} x \gt 0\\ \ln x \ne 0 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x \gt 0\\ x \ne 1 \end{array} \right..\]

Take the derivative using the quotient rule:

\[f^\prime\left( x \right) = \left( {\frac{x}{{\ln x}}} \right)^\prime = \frac{{x^\prime \cdot \ln x - x \cdot \left( {\ln x} \right)^\prime}}{{{{\ln }^2}x}} = \frac{{1 \cdot \ln x - x \cdot \frac{1}{x}}}{{{{\ln }^2}x}} = \frac{{\ln x - 1}}{{{{\ln }^2}x}}.\]

Equating the derivative to zero, we find the critical points \(c:\)

\[f^\prime\left( c \right) = 0,\;\; \Rightarrow \frac{{\ln c - 1}}{{{{\ln }^2}c}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {\ln c = 1}\\ {{{\ln }^2}c \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {c = e}\\ {c \ne 1} \end{array}} \right..\]

Note that the derivative does not exist at \(c = 1\) (where the denominator of the derivative approaches zero). But the function itself is also undefined at this point. Therefore, \(c = 1\) is not a critical point.

Hence, the function has one critical point \(c = e.\)

Example 9.

Determine critical points of the function \[f\left( x \right) = \left| {{x^2} - 4x + 3} \right|.\]

Solution.

First, we find the roots of the function and sketch its graph:

\[f\left( x \right) = 0,\;\; \Rightarrow \left| {{x^2} - 4x + 3} \right| = 0.\]
\[D = {\left( { - 4} \right)^2} - 4 \cdot 3 = 4,\;\; \Rightarrow {x_{1,2}} = \frac{{4 \pm \sqrt 4 }}{2} = 1,3.\]
Critical points of the function f(x)=|x^2-4x+3|
Figure 8.

We see that the function has two corner points (or V-points): \(c = 1\) and \(c = 3,\) where the derivative does not exist. Therefore, \(c = 1\) and \(c = 3\) are critical points of the function.

Besides that, the function has one more critical point at which the derivative is zero. Indeed, in the interval \(1 \le x \le 3,\) the function is written as

\[f\left( x \right) = - \left( {{x^2} - 4x + 3} \right) = - {x^2} + 4x - 3.\]

Differentiating and equating to zero, we get

\[f^\prime\left( x \right) = \left( { - {x^2} + 4x - 3} \right)^\prime = - 2x + 4.\]
\[f^\prime\left( c \right) = 0,\;\; \Rightarrow - 2c + 4 = 0,\;\; \Rightarrow c = 2.\]

Thus, the function has three critical points:

\[{{c_1} = 1,}\;{{c_2} = 2,}\;{{c_3} = 3.}\]

(a pretty nice answer, isn't it?)

Example 10.

Find the critical points of the function \[f\left( x \right) = {\sin ^2}x - \cos x\] on the interval \(\left( {0,2\pi } \right).\)

Solution.

Determine the derivative of \(f\left( x \right)\) using the chain rule and trig derivatives:

\[f^\prime\left( x \right) = \left( {{{\sin }^2}x - \cos x} \right)^\prime = 2\sin x\cos x - \left( { - \sin x} \right) = 2\sin x\cos x + \sin x = \sin x\left( {2\cos x + 1} \right).\]

Solving the equation \(f^\prime\left( c \right) = 0,\) we obtain two solutions:

\[f^\prime\left( c \right) = 0,\;\; \Rightarrow \sin c\left( {2\cos c + 1} \right) = 0.\]
\[1)\;\sin c = 0,\;\; \Rightarrow c = \pi n,\;n \in Z.\]

The equation \(\sin c = 0\) has one root \(c = \pi\) in the open interval \(\left( {0,2\pi } \right).\)

\[2)\;2\cos c + 1 = 0,\;\; \Rightarrow 2\cos x = - 1,\;\; \Rightarrow \cos c = - \frac{1}{2},\;\; \Rightarrow c = \pm \arccos \left( { - \frac{1}{2}} \right) + 2\pi n,\;\; \Rightarrow c = \pm \frac{{2\pi }}{3} + 2\pi n,\,n \in Z.\]

Only one solution \(c = \frac{{2\pi }}{3}\) belongs to the open interval \(\left( {0,2\pi } \right).\)

So, the function has two critical points:

\[{c_1} = \pi ,\;{c_2} = \frac{{2\pi }}{3}.\]

Example 11.

How many critical points does the function \[f\left( x \right) = \left| {{x^3} - 12x} \right|\] have?

Solution.

Critical points of the function f(x)=|x^3-12x|
Figure 9.

Determine the roots of the function:

\[f\left( x \right) = 0,\;\; \Rightarrow \left| {{x^3} - 12x} \right| = 0,\;\; \Rightarrow x\left( {{x^2} - 12} \right) = 0,\;\; \Rightarrow {x_1} = 0,\,x_{2,3} = \pm 2\sqrt 3 .\]

We see that the function has 3 corner points (or V-points) at \(x = - 2\sqrt 3 ,\) \(x = 0\) and \(x = 2\sqrt 3 .\) Since the derivative does not exist at these points, we have 3 critical points here.

Consider other critical points which can occur at local extrema.

In the interval \(\left[ { - 2\sqrt 3 ,0} \right],\) the function has the form

\[f\left( x \right) = {x^3} - 12x.\]

Its derivative is given by

\[f^\prime\left( x \right) = \left( {{x^3} - 12x} \right)^\prime = 3{x^2} - 12.\]

By equating the derivative to zero, we get

\[f^\prime\left( x \right) = 0,\;\; \Rightarrow 3{x^2} - 12 = 0,\;\; \Rightarrow x = \pm 2.\]

Only one point \(x = -2\) lies in the interval under consideration. So \(x = -2\) is also a critical point.

Similarly, we find that the function has one more critical point \(x = 2\) in the interval \(\left[ {0,2\sqrt 3 } \right]\).

Hence, the function has \(5\) critical points (\(3\) V-points and \(2\) local extrema points).

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